Question

I need some serious HELP!!!

Answer 1

ok, I'm using the integral formula for surface area to find the surface area of y=x^3, 0<x<2

Answer 2

here is what Ive got so far

Answer 3

\[2\pi \int\limits_{0}^{2} x (u)^{1/2} du/36x^3\]

Answer 4

so does the x^3 and the x completely cancel each other out

Answer 5

can you write the formula in general case ?!

Answer 6

Sorry, but we are not getting your question.
Could you please tell us which is the formula that you are using?
Surface area for rotation about any axis?

Answer 7

what do you mean in general case

Answer 8

The formula from which you started.

Answer 9

Ok, I dont think I said this. I'm looking for the surface area of y=x^3 if you were to rotate it about the x-axis. the interval is 0<x<2 does that help?

Answer 10

Yes! A sec please.

Answer 11

I think thats my fault I forgot to put where we were rotating said function

Answer 12

Thank you very much Im stuck at the integral part of the problem

Answer 13

\[S=2\pi \int\limits_{}^{}x^3\sqrt{1+(3x^2)^2}dx\]

Answer 14

right thats what i got except for the x^3 out front where did that come from?

Answer 15

how can I find the symbol "integral" in the eaquation editor?

Answer 16

3rd one from the top neemo

Answer 17

so where did the x^3 out in front come from?

Answer 18

So then:
\[S=2\pi \int\limits\limits_{0}^{2}x^3\sqrt{1+9x^4}dx=2\pi \int\limits_{}^{}\sqrt{u}du/36\]

Answer 19

The general formula is this one:

Answer 20

ok I was using the wrong one ok

Answer 21

\[S=2\pi \int\limits_{a}^{b}y \sqrt{1+f \prime}dx\]

Answer 22

Here you have what you need:
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
Find the sky blue section, on the middle of the page

Answer 23

@rafabc02 \[f \prime^2\] no ?!

Answer 24

Yes, you are right.

Answer 25

Thank you !

Answer 26

You are welcome!
@m_auld64
Could you get it?